Sorry but I could not resist:
alt text http://i43.tinypic.com/2v8k7lh.png
The map is defined on $[0,1]\times[0,1]$, and can be written as $$ f(x,y) = \begin{cases} (x, (2-x)y) & \text{ if }y\leq 1/2, \\ (x, xy+1-x) & \text{ if }y> 1/2. \end{cases} $$
Just take $D$ to be the set of $(x,y)$ with rational coordinates in $(0,1)\times(0,1)$, then $f$ is bijective on $D$ (because it can be easily inverted), but it is clearly not injective on $[0,1]\times[0,1]$.